bonanova
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Everything posted by bonanova
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Nope, just enjoy it a lot and try to go to great places. I have Scandinavian heritage and this was my first visit to Denmark [and Norway and Sweden]. Wonderful places.
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Wait ... there's more ... Enjoy!
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Bingo!
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Read one, two, three or all of these, depending on how desperate you feel.
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A fun place. I visited here last month.
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The question was how many rolls would it take for all numbers to show? The answer should be a number; your answer is a percentage. Are you saying that on 100% of the rolls of a fair die all six numbers show?
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You're on. I just did ten trials of four rolls. You got your 6 on four of the trials. I just repeated the ten trials. You got your 6 on five of them. That's 9 wins on 20 bets. I think that means I won. By the way, if you'd bet on 1 you'd have won 7 out of 20 times. if you'd bet on 2 you'd have won 7 out of 20 times. if you'd bet on 3 you'd have won 11 out of 20 times. if you'd bet on 4 you'd have won 10 out of 20 times. if you'd bet on 5 you'd have won 11 out of 20 times. On 17 of the 20 trials a number came up more than once. Keep in mind that if you bet on 6 and 6 comes up twice - say the four rolls give 5 3 6 6 - you get paid only once. But if a number you didn't bet on comes up more than once, your odds get longer.
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Totally agree that proofs are more useful than opinions. Let's do one. You refer to the following as a theorem of probability: If the probability of a favorable outcome after one trial is p, it will take on average 1/p trials to obtain it. Fair enough. Let's see if we can prove it, at least within the context of this puzzle. After showing say a '2' on the first roll we want a 1, 3, 4, 5 or 6 to show next. There are 6 possible outcomes; 5 are favorable. If we were to roll the die 6 times we would expect 5 favorable results [5 of 6 equally likely events are favorable]. Number of trials: n=6. Number of favorable outcomes: f=5. There are two things we can now say: The probability of a favorable outcome = f/n = 5/6 The number of trials for a favorable outcome = n/f = 6/5. The first quantity is p. The second quantity is 1/p. Thus, if the probability of a favorable result after one trial is p, then it takes 1/p trials to expect the favorable result.
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Not much action on this so let's close it out.
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OP = Original Post or Poster. A riddle is defined by the words used to pose it. A solution is an answer that fits the conditions of the riddle. Agree, if a hole had zero area, you wouldn't have to patch it.
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The OP says that a $5 patch fixes all the points on a closed hemisphere. Your case five points on the equator is a $5 fix.
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Agree this is a new riddle on BrainDen. As requested, here's a variant solution.
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Sure.
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Consider that the plane can be partitioned into unit squares. We could place a 0 in each square or we could place an 8 in each square. The squares have a 1-1 correspondence with the natural numbers. So, as Chuck Rampart points out, at least a countably infinite number of 0's and a countably infinite number of 8's can be placed in the plane. But is that it? Or could there be more? The title alludes to knowledge of how many 0's can be placed in the plane. Let's see what that is. Consider circles centered at the origin with radius r. If r is restricted to rational values [p/q where p and q are integers], this becomes another way to place a countably infinite number of 0's in the plane. But let r take on real number values, and we have uncountably infinite 0's. So now the question is whether you can do that with 8's. Can an uncountably infinite number of 8's be placed in the plane?
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Congrats for a $20 solution. Can the $20 strategy be worded to show it works for every case?
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